Question #05bee

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Answer 1 · 2016-08-09T15:18:14

#lim_(xrarr0)(sin(x^2-x))/x = -1# (I hope I've interpreted the question correctly.)

Use the fundamental trigonometric limit:

#lim_(theta rarr0) sin theta/theta = 1#

We'll make #theta = x^2-x# by multiplying by one to get #x^2-x# in the denominator.

#(sin(x^2-x))/x * (x-1)/(x-1) = (sin(x^2-x))/(x^2-x) * (x-1)#

Now the limit of the product is the product of the limits, so

#lim_(xrarr0)(sin(x^2-x))/x = lim_(xrarr0) (sin(x^2-x))/(x^2-x) * lim_(xrarr0)(x-1)#

# = (1)*(-1) = -1#